$A$ moving coil galvanometer has $48$ turns and the area of the coil is $4 \times 10^{-2} \, m^2$. If the magnetic field is $0.2 \, T$,then to increase the current sensitivity by $25\%$ without changing the area $(A)$ and the magnetic field $(B)$,the number of turns should become:

Two moving coil meters $M_1$ and $M_2$ have the following particulars:
$R_1 = 10\,\Omega, N_1 = 30, A_1 = 3.6 \times 10^{-3}\, m^2, B_1 = 0.25\, T$
$R_2 = 14\,\Omega, N_2 = 42, A_2 = 1.8 \times 10^{-3}\, m^2, B_2 = 0.50\, T$
(The spring constants are identical for the two meters). Determine the ratio of voltage sensitivity of $M_2$ and $M_1$.

In the circuit shown below,the ammeter and the voltmeter readings are $3 \text{ A}$ and $6 \text{ V}$ respectively. Then the value of the resistance $R$ is

The sensitivity of a galvanometer is $60 \text{ division/A}$. When a shunt is used,its sensitivity becomes $10 \text{ division/A}$. If the galvanometer is of resistance $20 \ \Omega$,the value of shunt used is (in $Omega$)

The resistance of a galvanometer is $25\,\Omega$ and it requires $50\,\mu A$ for full-scale deflection. The value of the shunt resistance required to convert it into an ammeter of $5\,A$ range is:

The magnets of two suspended coil galvanometers are of the same strength so that they produce identical uniform magnetic fields in the region of the coils. The coil of the first one is in the shape of a square of side $a$ and that of the second one is circular of radius $\frac{a}{\sqrt{\pi}}.$ When the same current is passed through the coils,the ratio of the torque experienced by the first coil to that experienced by the second one is